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Chapter 8: Problem 116

Do we lose our memory capacity as we get older? In a study of the effect of glucose on memory in elderly men and women, C. A. Manning and colleagues \(^{\star}\) tested 16 volunteers ( 5 men and 11 women) for long-term memory, recording the number of words recalled from a list read to each person. Each person was reminded of the words missed and was asked to recall as many words as possible from the original list. The mean and standard deviation of the long-term word memory scores were \(\bar{y}=79.47\) and \(s=25.25 .\) Give a \(99 \%\) confidence interval for the true long-term word memory scores for elderly men and women. Interpret this interval.

Short Answer

Expert verified
The 99% confidence interval is (60.87, 98.07).

Step by step solution

01

Identify the Given Information

We are given the sample mean \(\bar{y} = 79.47\), the sample standard deviation \(s = 25.25\), and the sample size \(n = 16\). We want to calculate a \(99\%\) confidence interval for the population mean.
02

Determine the Appropriate Formula

Since the population standard deviation is unknown and the sample size is small, we use the t-distribution to calculate the confidence interval. The formula for the confidence interval is given by:\[\bar{y} \pm t^* \cdot \left(\frac{s}{\sqrt{n}}\right)\]where \(t^*\) is the t-value corresponding to the \(99\%\) confidence interval with \(n-1\) degrees of freedom.
03

Find the Critical t-value

For a \(99\%\) confidence interval and \(15\) degrees of freedom (since \(n-1 = 15\)), we look up or calculate the critical t-value \(t^*\). Using a t-table or calculator, we find that \(t^* \approx 2.947\).
04

Calculate the Margin of Error

The margin of error is calculated as:\[E = t^* \cdot \left(\frac{s}{\sqrt{n}}\right) = 2.947 \cdot \left(\frac{25.25}{\sqrt{16}}\right)\]Simplifying, we find:\[E = 2.947 \cdot 6.3125 \approx 18.60\]
05

Calculate the Confidence Interval

Now, we calculate the confidence interval using the formula:\[(\bar{y} - E, \bar{y} + E)\]Substitute \(\bar{y} = 79.47\) and \(E \approx 18.60\):\[(79.47 - 18.60, 79.47 + 18.60) = (60.87, 98.07)\]
06

Interpret the Confidence Interval

We are \(99\%\) confident that the true mean of the long-term word memory scores for elderly men and women lies between \(60.87\) and \(98.07\). This interval provides a range of plausible values for the population mean memory score.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

t-distribution
When we want to calculate a confidence interval, especially with a small sample size, the t-distribution becomes quite handy. It helps us make inferences about the population mean when the population standard deviation is not known. Unlike the normal distribution that applies for larger sample sizes, the t-distribution has thicker tails. This means it accommodates the variability we have when dealing with small samples.
  • Applicability: Especially important for small samples (typically n < 30).
  • Degrees of freedom: Calculated as the sample size minus one, which affects the shape of the t-distribution, making it wider or narrower.
  • Flexibility: Adapts as sample size increases, eventually becoming similar to a normal distribution.
sample mean
The sample mean, denoted by \( \bar{y} \), is a measure of central tendency. It's the average of all observations in a sample and provides a reliable estimate of the center of a data set. By using the sample mean, we gain insights into the average performance or behavior within smaller sections of a population.
  • Calculation: Sum all observed values and divide by the number of observations in the sample.
  • Role in Confidence Intervals: Acts as the central point from which the interval is calculated.
  • Interpretation: Reflective of what might be expected in a broader context.
margin of error
The margin of error is an important part of a confidence interval. It indicates the range within which we expect the true population parameter to fall. Calculated as \( E = t^* \cdot \left( \frac{s}{\sqrt{n}} \right) \), it combines the critical t-value, standard deviation, and sample size to encapsulate variability.
  • Influenced by Sample Size: Larger sample sizes usually result in smaller margins, meaning more precise estimates.
  • Role of the Critical t-value: Affects the width, as higher confidence levels increase the margin of error.
  • Interpretation: Measures the uncertainty or possible error in the sample estimate.
critical t-value
The critical t-value, \( t^* \), is a pivotal element in determining the confidence interval. It is derived from the t-distribution and varies according to the desired confidence level and degrees of freedom. For a confidence level of 99%, the critical t-value is higher than it would be for, say, 95%, indicating a broader range.
  • Confidence Level Dependent: As confidence level increases, so does the critical t-value.
  • Degrees of Freedom: Calculated as the sample size minus one (n-1), influencing the t-value retrieved from tables.
  • Lookup: Can be found using statistical tables or calculators, necessary for precise interval estimation.

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Most popular questions from this chapter

The annual main stem growth, measured for a sample of 17 4-year-old red pine trees, produced a Thean of 11.3 inches and a standard deviation of 3.4 inches. Find a \(90 \%\) confidence interval for the mean annual main stem growth of a population of 4 -year-old red pine trees subjected to similar environmental conditions. Assume that the growth amounts are normally distributed.

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Suppose that two independent random samples of \(n_{1}\) and \(n_{2}\) observations are selected from normal populations. Further, assume that the populations possess a common variance \(\sigma^{2}\). Let $$S_{i}^{2}=\frac{\sum_{j=1}^{n_{i}}\left(Y_{i j}-\bar{Y}_{i}\right)^{2}}{n_{i}-1}, \quad i=1,2$$ a. Show that \(S_{p}^{2}\), the pooled estimator of \(\sigma^{2}\) (which follows), is unbiased: $$S_{p}^{2}=\frac{\left(n_{1}-1\right) S_{1}^{2}+\left(n_{2}-1\right) S_{2}^{2}}{n_{1}+n_{2}-2}$$ b. Find \(V\left(S_{p}^{2}\right)\)

When it comes to advertising, "tweens" are not ready for the hard-line messages that advertisers often use to reach teenagers. The Geppeto Group study \(^{\star}\) found that \(78 \%\) of 'tweens understand and enjoy ads that are silly in nature. Suppose that the study involved \(n=1030\) 'tweens. a. Construct a \(90 \%\) confidence interval for the proportion of 'tweens who understand and enjoy ads that are silly in nature. b. Do you think that "more than \(75 \%\) " of all 'tweens enjoy ads that are silly in nature? Why?

A pharmaceutical manufacturer purchases raw material from two different suppliers. The mean level of impurities is approximately the same for both suppliers, but the manufacturer is concerned about the variability in the amount of impurities from shipment to shipment. If the level of impurities tends to vary excessively for one source of supply, this could affect the quality of the final product. To compare the variation in percentage impurities for the two suppliers, the manufacturer selects ten shipments from each supplier and measures the percentage of impurities in each shipment. The sample variances were \(s_{1}^{2}=.273\) and \(s_{2}^{2}=.094\), respectively. Form a \(95 \%\) confidence interval for the ratio of the true population variances.
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